Let S be the set of all seven-digit numbers that can be formed using the digits 0, 1 and 2. For example, 2210222 is in S , but 0210222 is NOT in S . Then the number of elements x in S such that at least one of the digits 0 and 1 appears exactly t

Question

Let $S$ be the set of all seven-digit numbers that can be formed using the digits 0, 1 and 2. For example, 2210222 is in $S$ , but 0210222 is NOT in $S$ .

Then the number of elements $x$ in $S$ such that at least one of the digits 0 and 1 appears exactly twice in $x$ , is equal to ______. [2025]

Smart Achievers · Accepted Answer

(762)

Let $S$ “0” appear exactly twice, and $B \to$ “1” appear exactly twice.

$x$ $S$

$= \underset{(placing zero)}{{}^{6}C_{2}} \cdot (1) \cdot 2^{5} = \frac{6 \times 5}{2} \times 2^{5} = 480$

$For n (B)$

$Case 1: 1 at first place$

$Number of ways = \underset{(placing 1)}{{}^{6}C_{1}} \cdot (1) \cdot 2^{5} = 192$

$Case 2: 2 at first place$

$Number of ways = \underset{(placing 1)}{{}^{6}C_{2}} \cdot (1) \cdot 2^{4} = \frac{6 \times 5}{2} \times 2^{4} = 240$

$n (B) = 240 + 192$

$For n (A \cap B)$

$For n (A \cap B)$

$= \underset{placing zero}{{}^{6}C_{2}} \cdot (1) \times \underset{placing 1}{{}^{5}C_{2}} \cdot (1) \times \underset{2 at rest places 1}{(1)}$

$= \frac{6 \times 5}{2} \times \frac{5 \times 4}{2} = 150$

$∴$ $n (A \cup B) = n (A) + n (B) - n (A \cap B)$

$= 480 + (192 + 240) - 150 = 762$

Solution

Q.
Let $S$ be the set of all seven-digit numbers that can be formed using the digits 0, 1 and 2. For example, 2210222 is in $S$ , but 0210222 is NOT in $S$ .

Then the number of elements $x$ in $S$ such that at least one of the digits 0 and 1 appears exactly twice in $x$ , is equal to ______. [2025]

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Solution

Q. Let S be the set of all seven-digit numbers that can be formed using the digits 0, 1 and 2. For example, 2210222 is in S, but 0210222 is NOT in S. Then the number of elements x in S such that at least one of the digits 0 and 1 appears exactly twice in x, is equal to ______. [2025]

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